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tiny-mole [99]
4 years ago
9

Evaluate each expression. (1/3)^4

Mathematics
1 answer:
Leviafan [203]4 years ago
6 0

Answer:

1/81

Step-by-step explanation:

\huge \bigg( \frac{1}{3}  \bigg)^{4}  =  \frac{1}{ {3}^{4} }  =  \frac{1}{81}  \\

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Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com
Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.9842

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0
3 years ago
If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!
WITCHER [35]

Answer:

If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!If u do this u is smart and i also need help please report cards are next week!!!!!!!!!!!

Step-by-step explanation:

3 0
3 years ago
Classify the triangle.
adell [148]

Answer: Triangles can also be classified by their angles. In an acute triangle all three angles are acute (less than 90 degrees). A right triangle

Step-by-step explanation:Classification of triangles according to the length of their sides · Isosceles triangle · Equilateral triangle · Scalene Triangle.

3 0
3 years ago
An architect designed this scale model of a warehouse for a building contractor.
Annette [7]

tl;dr Answer is C

Here we will have to calculate 3 different areas separately.

When calculating the area of the triangle we will use the formula

A = (h*b)/2  

A = Area

h = height

b = base

To find the height we do X - Z  

23 - 15 = 8 ft

To find the base we do Y - W

19 - 13 = 6 ft

Using the formula above we can now solve for A  

A = (8*6)/2

A = (48)/2

A = 24 sq ft

Now we solve the two rectangles using the formula  

A = wl

w = width

l = length

We will calculate the area of the left most rectangle first.

We know the length of the rectangle because it's Y - W and we are given the width of the triangle.

w = 15 ft

l = 6 ft

A = 15*6

A = 90 sq ft

Second Rectangle has the width of X and length of W

w = 23 ft

l = 13 ft

A = 23 * 13

A = 299 sq ft

Now we add all the areas to give us the total area of the warehouse.

24 + 90 + 299 = 413 sq ft

Therefore, the answer is C

5 0
3 years ago
Can someone answer the blanks for points? THANKS I’m failing !
charle [14.2K]

Answer:

a. $3,333.33

b. $223,333.31

Step-by-step explanation:

a. $50,000 value increase and difference of 15 years 50,000/15 = 3,333.333333= 3333.33

b. difference 2006 and 1999 is 7 years, growth by year is $3333.33 so 3333.33 * 7 = 23,333.31 is difference in growth in those 7 years plus the original value of 1999 is $223,333.31

8 0
4 years ago
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